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Part BCSIR NET June 2024self-loop-needs-solving-not-summing

Self loop needs solving not summing

Let {XnX_{n} | n ≥ 0} be a homogeneous Markov chain with state space S = {0, 1, 2, 3, 4} and transition probability matrix

0 1 2 3 4
0 1/4 0 0 3/4 0
1 0 1 0 0 0
2 1/3 2/3 0 0 0
3 3/4 0 0 1/4 0
4 1/8 1/8 1/2 1/8 1/8

Let α\alpha denote the probability that starting with state 4 the chain will eventually get absorbed in closed class {0, 3}. Then the value of α\alpha is

  1. A.6/21
  2. B.11/21
  3. C.8/21
  4. D.10/21✓ Correct

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward, and why this one tests execution slip.

See pricing

50 are analysed free — try those first.

The trap it tests

Execution slip

The idea was right. The computation was not.

Drill statements like this

Related counterexample: An irreducible chain with a stationary distribution converges to it

More on this topic

The chapter behind this: Markov chains: classification and stationary behaviour — free to read

From Limit Theorems and Markov Chains › Markov chains: classification of states, stationary distributions

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